Gauss Law Problem With Conducting Charges?

Question

As a member of a team of storm physicists, you are attempting to replicate lightning by charging two long cables stretched over a canyon, as shown. One cable will attain a highly positive (and uniform) charge density of λ and the other will attain the same amount of charge density, but opposite in sign (i.e., –λ). Since the appearance of lightning directly depends on the electric field strength created by charge separation, it is important to derive an expression for electric field strength at all points between the two cables (albeit near the midpoint of the wires). The cables are sufficiently long as to be considered (for all practical purposes) infinitely long. Calculate the magnitude of the electric field strength between the two cables as a function of λ (the linear charge density) and r (the distance from the positively charged cable). Use ε0 as the permitivity of free space and assume the wires are separated by a distance D.

I cannot get this I keep getting the answer E=(Lambda)/(pi(epsilon)r)
Can some one please show me how to do this? Thank you.

Answer 1

The flux out of a Gaussian cylinder length L radius r containing a line charge density λ is:



∯E.dA = q_in/ε0 = λL/ε0



Since E is constant and radial over the cylinder (by symmetry) we can remove it from the integral:



E∯dA = λL/ε0



The surface area of the cylinder (excluding the ends where there is no flux) is:



∯dA = 2πrL



E1 = λ/2πε0r



For the negative charge density the field over a cylinder radius R is:



E2 = -λ/2πε0R



E at points between the two cables is given by the sum of the two fields. We may also use the fact that between the wires r + R = D to replace R in the equation for E2 giving the expression in terms of the radius from the positively charged cable:



E_tot = E1 + E2 = λ/2πε0r - λ/2πε0(D - r)



E_tot = (λ/2πε0)( (1/r) - (1/(D - r)) )

Answer 2

E dot ds = Int of [sum of enclosed charges (q) /eo]



Lambda = charge per unit area (L=1) in other words lambda = q/l = q

lamda = q

ds is the surface area of the wire can be thought of as a very long thin cylinder = 2 x pi x r x l

ds - 2 x pi x r x L



E (2 x pi x r x L) = lambda/ eo

E = Lambda/(2 x pi x r x L x eo)

Answer 3

The charged nsulating sphere induces and equivalent and opposite cost on the interior the around shell. evaluate a Gaussian floor mendacity fully interior the shell. The flux of the electrical powered container out of the Gaussian floor equals the full cost it encloses (Gauss' regulation) =0, considering the fact that this total cost incorporates the cost on the insulating sphere and the equivalent and opposite cost on the interior floor of the shell. that's conceivable provided that the sector interior the shell is 0. outdoors the shhere the sector is variety of a factor cost on the centre of magnitude the sum of the costs on the shell and insulating sphere.

Answer 4

E= \frac{\lambda }{2\pi \varepsilon_{0} } \left ( \frac{1}{r} +\frac{1}{D-r} \right )